Problem 14
Step-by-Step Solution
Verified Answer
The inequality holds by comparing AM and GM and using these bounds.
1Step 1: Analyze the Exercise
We need to prove the given inequality involving terms of arithmetic and geometric means of two numbers, \(a\) and \(b\), under the condition \(0 < a \leq b\). The inequality is:$$\frac{1}{8} \cdot \frac{(b-a)^{2}}{b} \leq \frac{a+b}{2}-\sqrt{a b} \leq \frac{1}{8} \cdot \frac{(b-a)^{2}}{a}$$
2Step 2: Establish the Mean Inequality
Recall the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), which states for two non-negative numbers, \(a\) and \(b\), that:\[\frac{a+b}{2} \geq \sqrt{a b}\]The difference between these two means can be expressed as \(\frac{a+b}{2} - \sqrt{a b}\). Our goal is to express this in a form that involves \((b-a)^2\).
3Step 3: Rewriting the Difference
Consider the expression:\[\left( \frac{a+b}{2} \right)^2 = \frac{a^2 + 2ab + b^2}{4}\]We want to compare this with \(\sqrt{a b}\). Now rewrite the square of the geometric mean:\[ ab = \sqrt{a b}^2 \]Calculate their difference:\[\left(\frac{a+b}{2}\right)^2 - ab = \frac{a^2 + 2ab + b^2}{4} - ab \]Simplifying this, we obtain:\[ \frac{a^2 - 2ab + b^2}{4} = \frac{(b-a)^2}{4}\]
4Step 4: Express the Difference of Means
Now, let's go back to:\[ \frac{a+b}{2} - \sqrt{a b} = \frac{\left( \frac{a+b}{2} \right)^2 - ab}{\frac{a+b}{2} + \sqrt{a b}} \]Substitute from Step 3:\[ = \frac{\frac{(b-a)^2}{4}}{\frac{a+b}{2} + \sqrt{a b}} \]Factor out this expression so we can compare with the given inequality.
5Step 5: Setting Up the Comparisons
With \(\frac{(b-a)^2}{4}\) already derived, we need to introduce factors so that the comparisons directly show:\[ \frac{(b-a)^2}{8b} \leq \frac{(b-a)^2}{4(a+b + 2\sqrt{a b})} \leq \frac{(b-a)^2}{8a} \]For the middle term, notice that \(a+b + 2\sqrt{a b}\) always lies between \(2a\) and \(2b\), justifying the inequality.
6Step 6: Finalize and Validate the Inequalities
Plugging everything back, confirm the bounds:- For the left inequality, since \(a + b + 2\sqrt{ab} \leq 4b\), it follows that \(\frac{(b-a)^2}{8b} \leq \frac{(b-a)^2}{4(a+b+2\sqrt{ab})}\).- For the right inequality, since \(a \leq a+b+2\sqrt{ab} \leq 4a\), it follows that \(\frac{(b-a)^2}{4(a+b+2\sqrt{ab})} \leq \frac{(b-a)^2}{8a}\).Thus, the inequalities are valid.
Key Concepts
Arithmetic MeanGeometric MeanMean InequalityInequality Proofs
Arithmetic Mean
The Arithmetic Mean (AM) is a fundamental mathematical concept often referred to as the average. It's calculated by summing a set of numbers and then dividing by the count of numbers. For example, the arithmetic mean of two numbers, \(a\) and \(b\), is given by the formula \(\frac{a+b}{2}\). This measure is useful because it represents a central point of the data set, offering a quick overview of the data's distribution.
In the context of AM-GM Inequality, the arithmetic mean of two non-negative numbers \(a\) and \(b\), reflects a balance between these numbers. Thus, it plays a crucial role in understanding how the numbers relate to each other in terms of their average spread.
Furthermore, because the Arithmetic Mean is often greater than or equal to other types of means, it is important when considering inequalities related to averages. This ties into inequality proofs where AM is usually a starting point for comparisons.
In the context of AM-GM Inequality, the arithmetic mean of two non-negative numbers \(a\) and \(b\), reflects a balance between these numbers. Thus, it plays a crucial role in understanding how the numbers relate to each other in terms of their average spread.
Furthermore, because the Arithmetic Mean is often greater than or equal to other types of means, it is important when considering inequalities related to averages. This ties into inequality proofs where AM is usually a starting point for comparisons.
Geometric Mean
The Geometric Mean (GM) is another type of mean, calculated differently than the arithmetic mean. For two numbers, \(a\) and \(b\), the geometric mean is \(\sqrt{ab}\). Unlike the arithmetic mean, the geometric mean considers the product of numbers rather than their sum. This makes it especially useful in situations where numbers are of different magnitudes or where multiplication and division are more relevant than addition and subtraction.
The geometric mean has a special property: it is always less than or equal to the arithmetic mean if the numbers involved are different. This is at the heart of the AM-GM Inequality. This property also allows for a balanced representation when dealing with ratios, such as in finance and sciences.
Understanding how the geometric mean interacts with other types of means, such as the arithmetic mean, is vital for proofs in inequality and for applications where multiplicative growth is discussed.
The geometric mean has a special property: it is always less than or equal to the arithmetic mean if the numbers involved are different. This is at the heart of the AM-GM Inequality. This property also allows for a balanced representation when dealing with ratios, such as in finance and sciences.
Understanding how the geometric mean interacts with other types of means, such as the arithmetic mean, is vital for proofs in inequality and for applications where multiplicative growth is discussed.
Mean Inequality
Mean inequalities, particularly the AM-GM Inequality, are foundational in mathematical analysis. The well-known AM-GM inequality states that for any non-negative real numbers \(a\) and \(b\), the arithmetic mean is greater than or equal to the geometric mean. Formally, \(\frac{a+b}{2} \geq \sqrt{ab}\).
This inequality is pivotal because it demonstrates a natural balance between additive and multiplicative factors of numbers. It suggests that a simple average (AM) of two numbers is at least as large as the somewhat weighted average GM for non-negative values. This inequality is employed in various proofs because it provides a clear criterion to establish relations between numbers when comparing means.
In inequality proofs, this principle can simplify complex problems, as it gives a direct way to reason about a relationship between sums and products. By providing bounds, it can help to estimate unknown components of a problem.
This inequality is pivotal because it demonstrates a natural balance between additive and multiplicative factors of numbers. It suggests that a simple average (AM) of two numbers is at least as large as the somewhat weighted average GM for non-negative values. This inequality is employed in various proofs because it provides a clear criterion to establish relations between numbers when comparing means.
In inequality proofs, this principle can simplify complex problems, as it gives a direct way to reason about a relationship between sums and products. By providing bounds, it can help to estimate unknown components of a problem.
Inequality Proofs
Inequality proofs are a large and important part of higher mathematics. They involve showing that one expression is greater than or less than another under certain conditions. The process is systematic and relies on known inequalities such as the AM-GM Inequality.
In the given exercise, the inequality proof requires showing the bounds for \(\frac{a+b}{2} - \sqrt{ab}\). The approach involves manipulating both sides of the inequality until you establish the given relationships for the differences in means. Such proofs typically include:
Successful inequality proofs often involve simplifying expressions and finding relationships that aren't immediately obvious, serving as tools for problem-solving across different areas of mathematics.
In the given exercise, the inequality proof requires showing the bounds for \(\frac{a+b}{2} - \sqrt{ab}\). The approach involves manipulating both sides of the inequality until you establish the given relationships for the differences in means. Such proofs typically include:
- Identifying expressions that need bounding;
- Using known inequalities, like AM-GM, to guide the transformations;
- Rewriting expressions to clearly reveal the bounds.
Successful inequality proofs often involve simplifying expressions and finding relationships that aren't immediately obvious, serving as tools for problem-solving across different areas of mathematics.
Other exercises in this chapter
Problem 13
Find all functions \(\boldsymbol{f}: \mathrm{R} \backslash\\{-1,0\\} \rightarrow \mathbb{R}\) such that $$ x \in \mathbb{R} \backslash(-1,0\\} \Longrightarrow f
View solution Problem 13
Let \(A, B\), and \(C\) be sets. Shew that $$ A \times(B \backslash C)=(A \times B) \backslash(A \times C) $$
View solution Problem 14
Let \(f^{[1]}(x)=f(x)=2 x, f^{[n+1]}=f \circ f^{[n]}, n \geq 1\) Find a closed formula for \(\boldsymbol{f}^{[\boldsymbol{n} \mid}\)
View solution Problem 14
Prove that a set with \(N \in \mathbb{N}\) elements has exactly \(2^{N}\) subsets.
View solution